%I #18 Jan 17 2018 03:23:46

%S 0,1,3,7,30,144,876,6084,48816,438624,4389120,48263040,579242880,

%T 7529552640,105417365760,1581231456000,25299906508800,430096581734400,

%U 7741753102540800,147093162635059200,2941864569520128000

%N Numerators of unsigned columns of triangle A110504: a(n) = n!*A110504(n,0) = (-1)^k*n!*A110504(n+k,k) for all k >= 0.

%C Triangle A110504 equals the matrix logarithm of triangle A110503.

%C Triangle A110503 shifts one column left under matrix inverse.

%C Lim_{n->infinity} a(n)/n! = Pi*2*sqrt(3)/9 = 1.209199576...

%H Vaclav Kotesovec, <a href="/A110505/b110505.txt">Table of n, a(n) for n = 0..440</a>

%F E.g.f.: (2+x-x^2)/(2*(1-x)) * arccos(1-x^2/2) / sqrt(1-x^2/4).

%F E.g.f. A(x) satisfies:

%F (1) A(x)*A(-x) = -arccos(1-1/2*x^2)^2 = Sum_{n>=0} -x^(2*n+2)/( C(2*n+1, n)*(n+1)^2 ).

%F (2) 1/(1-x) = Sum_{n>=1} A(x)^floor((n+1)/2) * A(-x)^floor(n/2)/n!.

%F a(2*n+1) = (2*n+1)!*(1 + Sum_{k=1..n} (1/binomial(2*k+1, k))/(k+1)).

%F a(2*n+2) = (2*n+2)!*(1 + 1/2 - Sum_{k=1..n} 1/binomial(2*k+2, k)/k) = n!*(1 + 1/2 - 1/3 + 1/12 - 1/20 + 1/60 - 1/105 + 1/280 -+ ...).

%F Recurrence: 4*a(n) = 2*(2*n-1)*a(n-1) + (n-2)*(n+1)*a(n-2) - (n-3)*(n-2)*n*a(n-3). - _Vaclav Kotesovec_, May 09 2014

%e E.g.f.: A(x) = x + 3*x^2/2! + 7*x^3/3! + 30*x^4/4! + 144*x^5/5! + 876*x^6/6! + ...

%e where A(x) satisfies: A(x)*A(-x) = -arccos(1-1/2*x^2)^2, and

%e arccos(1-1/2*x^2)^2 = Sum_{n>=0} x^(2*n+2)/( C(2*n+1, n)*(n+1)^2 ) = x^2 + 1/12*x^4 + 1/90*x^6 + 1/560*x^8 + 1/3150*x^10 + ...

%o (PARI) /* From relation to unsigned columns of triangle A110504: */

%o {a(n)=local(M=matrix(n+1,n+1,r,c,if(r>=c, if(r==c || c%2==1,1,if(r%2==0 && r==c+2,-2,-1))))); n!*sum(i=1,#M,-(M^0-M)^i/i)[n+1,1]}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) /* As Partial Sums of Series: */

%o a(n)=if(n<1,0,n!*(1+sum(n=2,n,(-1)^n/(binomial(n-2,n\2-1)*n*(n-1)/((n+1)\2)))))

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A110503 (triangular matrix), A110504 (matrix logarithm), A002544.

%K frac,nonn

%O 0,3

%A _Paul D. Hanna_, Jul 23 2005